Prove that there is a positive integer that equals the

berljivx8

berljivx8

Answered question

2021-12-17

Prove that there is a positive integer that equals the sum of the positive integers not exceeding it. Is your proof constructive or nonconstructive?

Answer & Explanation

enlacamig

enlacamig

Beginner2021-12-18Added 30 answers

Assume n is a positive integer that equals the sum of the positive integers not exceeding it:
n=n+(n1)+(n2)+(n3)++2+1
Using summation formula we get:
n=n(n+1)22n=n2+n0=n2n0=n(n1)
Therefore, n=0 or n=1. But n is a nonzero integer, so it must be n=1.
Result: We prove that there exists such positive integer and it is n=1, so our proof is constructive
nick1337

nick1337

Expert2023-05-27Added 777 answers

Let n be a positive integer. We want to show that there exists a positive integer k such that k=i=1ni.
By the formula for the sum of an arithmetic series, we have:
i=1ni=n(n+1)2.
Now, consider the positive integer k=n(n+1)2. We need to show that this value of k satisfies the required condition.
Substituting k into the sum expression, we have:
k=i=1ni=n(n+1)2.
Hence, we have proven that there exists a positive integer k (specifically, k=n(n+1)2) that equals the sum of the positive integers not exceeding it.
Don Sumner

Don Sumner

Skilled2023-05-27Added 184 answers

Step 1:
To prove that there is a positive integer that equals the sum of the positive integers not exceeding it, we can use a nonconstructive proof.
Let n be the positive integer we are considering. The sum of the positive integers not exceeding n can be expressed as 1+2+3++n, which is the sum of an arithmetic series. Using the formula for the sum of an arithmetic series, we have:
n(n+1)2
Step 2:
Now, we need to show that there exists a positive integer n such that n(n+1)2=n. This can be rewritten as:
n(n+1)=2n
Expanding the equation, we get:
n2+n=2n
Simplifying further:
n2n=0
Factoring out n, we have:
n(n1)=0
This equation holds true for n=0 and n=1. Therefore, there exists a positive integer (n=1) that satisfies the equation.
Vasquez

Vasquez

Expert2023-05-27Added 669 answers

1. Assume n is the positive integer we seek.
2. The sum of positive integers not exceeding n can be expressed as k=1nk.
3. By using the formula for the sum of an arithmetic series, we can rewrite the sum as n(n+1)2.
4. Now, we need to find an n that satisfies the equation n(n+1)2=n.
5. Simplifying the equation, we have n2+n2=n.
6. Multiplying both sides of the equation by 2, we get n2+n=2n.
7. Rearranging the terms, we have n2n=0.
8. Factoring out n, we get n(n1)=0.
9. Therefore, the equation is satisfied when n=0 or n=1.
10. However, since we are looking for a positive integer, we discard n=0.
11. Thus, the positive integer that equals the sum of the positive integers not exceeding it is n=1.
Therefore, the proof is constructive, as we have explicitly found the positive integer that satisfies the given condition, which is n=1.

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